A 40 cm wire having a mass of  3.2 g  is stretched between two fixed supports 40.05 cm apart. In its fundamental mode, the wire vibrates at 220 Hz.   If the area of cross-section of the wire is $1.0m{m}^2$ , then young’s modulus of the wire is $\_\_\_\_\times {10}^{11}N/m^2$

$$\begin{gathered} Given : L=40c.m , m=3.2g , \Delta L=40.05-40=0.05c.m , f_1=220Hz , A=1m{m}^2 \\ Young\text{'}s Modulud : Y = \frac{{\displaystyle \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$A$}\right.}}{{\displaystyle \raisebox{1ex}{$∆L$}\!\left/ \!\raisebox{-1ex}{$L$}\right.}} \\ \Rightarrow T=AY\frac{\Delta L}{L}=\frac{AY\left(0.05\right)}{40}=\frac{AY}{800} \\ \Rightarrow T=\frac{AY}{800}=\frac{{10}^{-6}\times Y}{800} \\ Linear Mass Density : \mu =m/L=\frac{3.2\times {10}^{-3}}{40\times {10}^{-2}}=8\times {10}^{-3}kg/m \\ Fundamental Frequency : f_1=\frac{1}{2L}\sqrt{\frac{T}{\mu }} \\ \Rightarrow 220=\frac{1}{2\times 0.4}\sqrt{\frac{{10}^{-6}\times Y}{800\times 8\times {10}^{-3}}} \\ \Rightarrow 176=\sqrt{\frac{Y}{6.4\times {10}^6}} \\ \Rightarrow Y=6.4\times {\left(176\right)}^2\times {10}^6=1.98\times {10}^{11}N/m^2 \end{gathered}$$